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a)\(\dfrac{3}{10}\)-x=\(\dfrac{25}{30}\)-\(\dfrac{4}{30}\)
\(\dfrac{3}{10}-x=\dfrac{7}{10}\)
x = \(\dfrac{3}{10}-\dfrac{7}{10}\)
x=\(\dfrac{-4}{10}\)
b)\(\dfrac{-5}{8}+x=\dfrac{4}{9}-\dfrac{63}{9}\)
\(\dfrac{-5}{9}+x=\dfrac{-59}{9}\)
\(x=\dfrac{-59}{9}-\dfrac{-5}{9}\)
\(x=\dfrac{-64}{9}\)
c)=>2.18=(x-3).(x-3)
=>36=(x-3)\(^2\)
=>6\(^2\)=(x-3)\(^2\)
6= x-3
x=6+3=9
9: \(=1-\dfrac{1}{99}+1-\dfrac{1}{100}+\dfrac{100}{101}\cdot\dfrac{1-4+3}{12}=2-\dfrac{199}{9900}=\dfrac{19601}{9900}\)
10: \(=\left(\dfrac{78}{79}+\dfrac{79}{80}+\dfrac{80}{81}\right)\cdot\dfrac{6+5+9-20}{30}=0\)
a) 80<243 nên \(\frac{1}{80}>\frac{1}{243}\)
\(\Rightarrow\left(\frac{1}{80}\right)^7>\left(\frac{1}{243}\right)^7\) mà \(\left(\frac{1}{243}\right)^7>\left(\frac{1}{243}\right)^6\)
Nên \(\left(\frac{1}{80}\right)^7>\left(\frac{1}{243}\right)^6\)
b) Ta so sánh \(\frac{3}{8}\) với \(\frac{5}{243}\)
Ta có: \(\frac{3}{8}=\frac{3.243}{8.243}=\frac{729}{1944}\)
\(\frac{5}{243}=\frac{5.8}{243.8}=\frac{40}{1944}\)
Suy ra: \(\frac{3}{8}>\frac{5}{243}\)
\(\Rightarrow\left(\frac{3}{8}\right)^5>\left(\frac{5}{243}\right)^5\) mà \(\left(\frac{5}{243}\right)^5>\left(\frac{5}{243}\right)^3\)
Nên \(\left(\frac{3}{8}\right)^5>\left(\frac{5}{243}\right)^3\)
A>B
bạn ơi tính ra giúp mk đc ko???